802
11.7 Nonhomogeneous Linear Systems
Variation of Parameters
The method of variation of parameters is a general method for
solving a linear nonhomogeneous system
x0 = Ax + F(t).
Historically, it was a trial solution method, whereby the nonhomogeneous
system is solved using a trial solution of the form
x(t) = eAt x0 (t).
In this formula, x0 (t) is a vector function to be determined. The method
is imagined to originate by varying x0 in the general solution x(t) =
eAt x0 of the linear homogenous system x0 = Ax. Hence was coined the
names variation of parameters and variation of constants.
Modern use of variation of parameters is through a formula, memorized
for routine use.
Theorem 28 (Variation of Parameters for Systems)
Let A be a constant n × n matrix and F(t) a continuous function near
t = t0 . The unique solution x(t) of the matrix initial value problem
x0 (t) = Ax(t) + F(t),
x(t0 ) = x0 ,
is given by the variation of parameters formula
(1)
x(t) = eAt x0 + eAt
Z t
e−rA F(r)dr.
t0
Proof of (1). Define
Z
t
u(t) = x0 +
e−rA F(r)dr.
t0
To show (1) holds, we must verify x(t) = eAt u(t). First, the function u(t) is
differentiable with continuous derivative e−tA F(t), by the fundamental theorem
of calculus applied to each of its components. The product rule of calculus
applies to give
0
x0 (t) = eAt u(t) + eAt u0 (t)
= AeAt u(t) + eAt e−At F(t)
= Ax(t) + F(t).
Therefore, x(t) satisfies the differential equation x0 = Ax + F(t). Because
u(t0 ) = x0 , then x(t0 ) = x0 , which shows the initial condition is also satisfied.
The proof is complete.
11.7 Nonhomogeneous Linear Systems
803
Undetermined Coefficients
The trial solution method known as the method of undetermined coefficients can be applied to vector-matrix systems x0 = Ax + F(t) when the
components of F are sums of terms of the form
(polynomial in t)eat (cos(bt) or sin(bt)).
Such terms are known as atoms. It is usually efficient to write F in
terms of the columns e1 , . . . , en of the n × n identity matrix I, as the
combination
n
F(t) =
X
Fj (t)ej .
j=1
Then
x(t) =
n
X
xj (t),
j=1
where xj (t) is a particular solution of the simpler equation
x0 (t) = Ax(t) + f (t)c,
f = Fj ,
c = ej .
An initial trial solution x(t) for x0 (t) = Ax(t) + f (t)c can be determined
from the following initial trial solution rule:
Assume f (t) is a sum of atoms. Identify independent functions whose linear combinations give all derivatives of f (t).
Let the initial trial solution be a linear combination of these
functions with undetermined vector coefficients {cj }.
In the well-known scalar case, the trial solution must be modified if its
terms contain any portion of the general solution to the homogeneous
equation. If f (t) is a polynomial, then the correction rule for the initial
trial solution is avoided by assuming the matrix A is invertible. This
assumption means that r = 0 is not a root of det(A − rI) = 0, which
prevents the homogenous solution from having any polynomial terms.
The initial vector trial solution is substituted into the differential equation to find the undetermined coefficients {cj }, hence finding a particular
solution.
Theorem 29 (Polynomial solutions)
P
j
Let f (t) = kj=0 pj tj! be a polynomial of degree k. Assume A is an n × n
constant invertible matrix. Then u0 = Au + f (t)c has a polynomial solution
P
j
u(t) = kj=0 cj tj! of degree k with vector coefficients {cj } given by the
relations
cj = −
k
X
i=j
pi Aj−i−1 c,
0 ≤ j ≤ k.
804
Theorem 30 (Polynomial × exponential solutions)
P
j
Let g(t) = kj=0 pj tj! be a polynomial of degree k. Assume A is an n × n
constant matrix and B = A − aI is invertible. Then u0 = Au + eat g(t)c
P
j
has a polynomial-exponential solution u(t) = eat kj=0 cj tj! with vector
coefficients {cj } given by the relations
cj = −
k
X
pi B j−i−1 c,
0 ≤ j ≤ k.
i=j
Proof of Theorem 29. Substitute u(t) =
Pk
j=0
j
cj tj! into the differential
equation, then
k−1
X
cj+1
j=0
k
k
X
tj X tj
tj
=A
cj +
pj c.
j!
j! j=0 j!
j=0
Then terms on the right for j = k must add to zero and the others match the
left side coefficients of tj /j!, giving the relations
Ack + pk c = 0,
cj+1 = Acj + pj c.
Solving these relations recursively gives the formulas
ck
ck−1
= −pk A−1 c,
= − pk−1 A−1 + pk A−2 c,
..
.
c0
= − p0 A−1 + · · · + pk A−k−1 c.
The relations above can be summarized by the formula
cj = −
k
X
pi Aj−i−1 c,
0 ≤ j ≤ k.
i=j
Pk
tj
The calculation shows that if u(t) =
j=0 cj j! and cj is given by the last
formula, then u(t) substituted into the differential equation gives matching
LHS and RHS. The proof is complete.
Proof of Theorem 30. Let u(t) = eat v(t). Then u0 = Au + eat g(t)c implies
v0 = (A − aI)v + g(t)c. Apply Theorem 29 to v0 = Bv + g(t)c. The proof is
complete.